Quaternion Exponential Map - Lie group vs. Riemannian Manifold I have a number of related questions on how the quaternion exponential map can be defined, and in trying to read up on the topic I've gone down a rabbit hole in fields I have no background in. I apologize in advance if my terminology is off here; I'll try to summarize what I know to clarify my questions.


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*Do both the set of imaginary quaternions and the set of unit quaternions parameterize $SU(2)$ via the exponential map? The classical axis-angle mapping shows that the unit quaternions do, so that seems sufficient proof in that case. For the imaginary (but not necessarily unit) case, I think that the exponential map should still work since the scaling of the quaternion relative to a unit pure quaternion provides the same information as a unit quaternion with nonzero real part.

*Is it true that each of the following three sets constitutes a Lie group: The set of unit quaternions, the set of imaginary quaternions, and the set of all nonzero quaternions?

*According to Wikipedia, the exponential map defined as above should coincide with the exponential map defined for the Riemannian manifold of the quaternions, but it's not clear to me how to define the appropriate metric for this to be the case. For the unit quaternions, the tangent space at 1 is just the imaginary quaternions; for non-unit quaternions, isn't the tangent space the same? But in that case, since the definition of the Lie exponential map says that the exponential of a quaternion is defined according to a subgroup whose tangent vector at the identity is q, if q is not pure imaginary then how do we define the exponential? I'm generally confused about how we can see the correspondence between the Lie and Riemannian exponential maps for quaternions. Is the tangent vector in this case just some scaled version of the imaginary part of the quaternion?

*Finally, if we can define the Riemannian exponential map as above to coincide with the Lie exponential map, the formula for quaternions that I have seen suggests it should be defined as $q*exp(d)$. What does this mapping mean, and how does it relate to the Riemannian exponential map? Also, is this definition still meaningful for non-unit quaternions?

 A: (Personally I use $\mathrm{SU}(2)$ strictly to denote the group of $2\times 2$ special unitary matrices and $\mathrm{Sp}(1)$ to denote the set of unit quaternions, but most people use them interchangeably since they are isomorphic Lie groups, like $\mathrm{U}(1)$ and $\mathrm{SO}(2)$.)

Do both the set of imaginary quaternions and the set of unit quaternions parameterize SU(2) via the exponential map?

This is kind of like asking "do both pairs of angles and the sphere parametrize  the sphere via sines and cosines?" It's a weird thing to say. One uses sines and cosines to parametrize the sphere using two angles, sure, but one uses ... the identity function to parametrize a space with itself.
The unit quaternions are $\mathrm{Sp}(1)$, and each unit quaternion has a polar form given by the formula$\exp(\theta\mathbf{u})=\cos(\theta)+\sin(\theta)\mathbf{u}\,$ (just like complex numbers) where $\mathbf{u}$ is unit vector. Thus, the exponential function maps $\mathrm{Im}(\mathbb{H})=\mathbb{R}^3$ to the group $\mathrm{Sp}(1)$.
I don't understand the rest of what you're saying in question (1).

Is it true that each of the following three sets constitutes a Lie group: The set of unit quaternions, the set of imaginary quaternions, and the set of all nonzero quaternions?

The unit quaternions and nonzero quaternions are multiplicative groups and the imaginary quaternions are a vector space under addition, but the imaginary quaternions are certainly not closed under multiplication. Given two vectors $\mathbf{u}$ and $\mathbf{v}$, the product's real and imaginary parts (i.e. its scalar and vector components) are given by the formula $\mathbf{uv}=-\mathbf{u}\cdot\mathbf{v}+\mathbf{u}\times\mathbf{v}$, so the (quaternion) product of two vectors will itself be a vector if and only if they're orthogonal.

It's not clear to me how to define the appropriate metric for this to be the case.

There's already a metric on $\mathbb{H}$. It's given by
$$ |x_0+x_1\mathbf{i}+x_2\mathbf{j}+x_3\mathbf{k}|=\sqrt{x_0^2+x_1^2+x_2^2+x_3^2}. $$
So if you have a nice curve $\gamma:[0,1]\to\mathbb{H}$ its length can be calculated by $\int_0^1 |\gamma'(t)|\,\mathrm{d}t$, just as in any Euclidean space $\mathbb{R}^n$. Indeed, just as in $\mathbb{R}^2$ and $\mathbb{R}^3$, the length of an arc between two points on the unit sphere will be the angle between them.

For the unit quaternions, the tangent space at $1$ is just the imaginary quaternions; for non-unit quaternions, isn't the tangent space the same?

Nope. For inspiration, just look at $S^2$. Clearly antipodal points have the same tangent plane (interpreted as vector subspaces of $\mathbb{R}^3$ anyway), but otherwise the various tangent planes are not parallel and exist in different orientations.
Suppose $q:[0,1]\to\mathrm{Sp}(1)$ is a differentiable curve in the unit quaternions, and that $q(0)=1$. We have $q(t)\overline{q(t)}=1$; differentiating and then evaluating at $t=0$ yields $q'(0)+\overline{q'(0)}=0$, which is only possible if $q'(0)$ is purely imaginary since $x+\overline{x}=2\mathrm{Re}(x)$. But this argument relies on $q(0)=1$, i.e. on the fact we're look at the tangent space at $q=1$.
In general if $q(0)=q$ then instead you get $q'(0)\overline{q}+q\overline{q'(0)}=0$, or equivalently $2\mathrm{Re}(q'(0)\overline{q})=0$, so that $q'(0)\overline{q}$ is pure imaginary, or in other words $q'(0)$ is an an imaginary quaternion multiplied by $q$ from the right. Since $q^{-1}\mathbf{x}q$ is just $\mathbf{x}$ rotates, and $\mathbf{x}q=q(q^{-1}\mathbf{x}q)$, we could have equally well have said multiplying by $q$ from the left instead of the right.
In conclusion, the act of multiplying by $q$ slides (is a map from) the tangent space $T_1S^3$ to the tangent space $T_qS^3$. (Note $S^3$ is another term for $\mathrm{Sp}(1)$, just as $S^1$ is notation for $\mathrm{U}(1)$.)
If you want the exponential map $\exp:T_qS^3\to S^3$ in the Riemannian manifold sense, you can do a "transport of structure" argument: slide $T_qS^3$ to $T_1S^3$, apply the usual $\exp$ map, then slide back (so that $\exp$ of the zero vector in $T_qS^3$ yields $q$). Depending on if you slide using left or right multiplications, this yields $q\exp(\overline{q}x)$ or $\exp(x\overline{q})q$ (which are equal to each other).
Hopefully this answers (4) as well for you.
